If the first and the last terms of an arithmetic series are 10 and 62, show that the sum of the series varies directly as the number of terms
a=10
a+(n-1)d = 62
10 + nd - d = 62
nd-d=52
n=(52+d)/d or d = 52/(n-1)
sum(n) = (n/2)(2a + (n-1)d)
= (n/2)(20 + (n-1)(52)/(n-1))
= (n/2)(20 + 52)
= (n/2)(72) = 36n
sum(n) = 36n
so sum(n) varies directly as the number of terms
To show that the sum of the arithmetic series varies directly as the number of terms, we need to express the sum of the series as a function of the number of terms and then show that the function represents a direct variation relationship.
Let's start by finding the common difference (d) of the arithmetic series. The common difference is the difference between consecutive terms. We can calculate it using the formula:
d = (last term - first term) / (number of terms - 1)
Given that the first term (a₁) is 10, the last term (aₙ) is 62, and the number of terms (n) is unknown, we can substitute these values into the formula:
d = (62 - 10) / (n - 1)
d = 52 / (n - 1)
Now, we can express the sum of the arithmetic series (Sₙ) as a function of the number of terms (n) using the formula:
Sₙ = (n/2) * (a₁ + aₙ)
Substituting the given values, we have:
Sₙ = (n/2) * (10 + 62)
Sₙ = 36n
Therefore, the sum of the series (Sₙ) varies directly as the number of terms (n), with the constant of variation being 36.
In conclusion, we followed the steps of finding the common difference, expressing the sum of the series as a function of the number of terms, and verifying that it represents a direct variation relationship with a constant of variation.